Question

Three bulbs are connected by tubing, and the tubing is evacuated. The volume of the tubing is 37.0 mL. The first bulb has a volume of 89.0 mL and contains 7.92 atm of argon, the second bulb has a volume of 250 mL and contains 1.94 atm of neon, and the third bulb has a volume of 38.0 mL and contains 3.42 atm of hydrogen. If the stopcocks (valves) that isolate all three bulbs are opened, what is the final pressure of the whole system in atm?

Answer 1

Explanation:

Formula to calculate final pressure is as follows.

       $P_{final} \times V_{final}  = P_{1}V_{1} + P_{2}V_{2} + P_{3}V_{3} + P_{4}V_{4}$ ............... (1)

Since, earlier the tube had no gas therefore, its pressure was zero initially considering temperature remains constant.

So, total number of moles will be as follows.

         n = $n_{1} + n_{2} + n_{2}$

As, T and R will be common on both the sides. Hence, they will cancel out from both the sides.

Therefore, putting the given values into equation (1) as follows.

            $89 ml \times 7.92 + 250 ml \times 1.94 atm + 38.0 ml \times 3.42 atm = P final \times (37 ml + 89.0 ml + 250 ml + 38.0 ml)$

           704.88 + 485.0 + 129.96 = $P_{final} \times 414$

                   $P_{final} = \frac{1319.84}{414}$

                                   = 3.18 atm

Therefore, we can conclude that final pressure of the whole system in atm is 3.18 atm.